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RD Sharma XII Vol 1 2020 Solutions for Class 12 Humanities Maths Chapter 9 Differentiability are provided here with simple step-by-step explanations. These solutions for Differentiability are extremely popular among class 12 Humanities students for Maths Differentiability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 1 2020 Book of class 12 Humanities Maths Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 1 2020 Solutions. All RD Sharma XII Vol 1 2020 Solutions for class 12 Humanities Maths are prepared by experts and are 100% accurate.

Page No 9.10:

Question 1:

Show that f(x) = |x − 2| is continuous but not differentiable at x = 2.

Answer:

Given: f(x) = |x-2| = x-2,     x≥2-x+2,   x<2
 
Continuity at x=2: We have,

(LHL at x = 2)
 =limx→2- f(x) = limh→0 f(2-h) = lim h→0  (-2+h)+2= 0.

(RHL at x = 2)
  =lim     x→2+f(x) = limh→0 f(2+h) = limh→0  2+h-2 = 0.

and f(2) = 0

Thus, limx→2- f(x) = limx→2+ f(x) = f(2)f(2).
Hence, f(x) is continuous at x=2.

 Differentiability at x = 2: We have,

(LHD at x = 2)
 =limx→2- f(x) - f(2)x-2 = limx→2 (-x+2) - 0x-2 = limx→2 -(x-2)x-2 = limx→2 (-1) =-1

(RHD at x=2)
= =limx→2+ f(x) - f(2)x-2 = limx→2 (x-2) - 0x-2 = limx→2 1 = 1

Thus, limx→2- f(x) ≠ limx→2+ f(x).

Hence, f(x) is not differentiable at x=2 .

Page No 9.10:

Question 2:

Show that f(x) = x1/3 is not differentiable at x = 0.

Answer:

Disclaimer: It might be a wrong question because f(x) is differentiable at x=0

Given: f(x) = x13.
We have,
(LHD at x = 0)
 limx→0- f(x) - f(0)x-0=limh→0 f(0-h) - f(0)0-h-0=limh→0 0-h13-013-h=limh→0 -h13-h=limh→0  -h-23= 0
           
(RHD at x = 0)
 limx→0+ f(x) - f(0)x-0=limh→0 f(0+h) - f(0)0+h-0=limh→0 0+h13-013-h=limh→0 h13h=limh→0  h-23= 0                      

LHD at (x = 0)= RHD at (x = 0)

Hence,  f(x) = x13 is differentiable at x = 0

Page No 9.10:

Question 3:

Show that fx=12 x-13, if x≤32x2+5, if x>3 is differentiable at x = 3. Also, find f'(3).

Answer:

Given: f(x) =12x-13,  x≤3.2x2 + 5,   x>3.

We have to show that the given function is differentiable at x = 3.

We have,

(LHD at x=3) = lim x→3- f(x) - f(3)x-3
 
                      = lim x→3 12x-13 - 23x-3= limx→3  12x-36x-3= limx→3  12 (x-3)x-3= limx→3 12 = 12

(RHD at x = 3) = limx→3+ f(x) - f(3)x-3

                        = limx→3 2x2+5 - 23x-3= limx→3 2x2 -18x-3= limx→3 2 (x2 -9)x-3=limx→3 2(x+3) = 2×6 = 12

Thus, (LHD at x=3) = (RHD at x=3) = 12.

So, f(x) is differentiable at x=3 and f'(3) = 12.

Page No 9.10:

Question 4:

Show that the function f defined as follows, is continuous at x = 2, but not differentiable thereat: fx=3x-2,0<x≤12x2-x,1<x≤25x-4,x>2

Answer:

Given: 
         f(x) =  3x-2,    0<x≤12x2-x,  1<x≤25x-4,     x>2

First , we will show that f(x) is continuos at x=2.

We have,

(LHL at x=2)

=lim      x→2- f(x) = lim h→0 f(2-h)  = limh→0 2(2-h)2 - (2-h) = limh→0 (8 + 2h2 - 8h - 2 + h) = 6

(RHL at x = 2) 

=limx→2+ f(x) = limh→0  f(2+h) = limh→0 5(2+h) - 4 = limh→0 (10 + 5h -4) = 6

and f(2) = 2×4 - 2 = 6.           

Thus,  limx→2- f(x) = limx→2+ f(x) = f(2).

Hence the function is continuous at x=2.


Now, we will check whether the given function is differentiable at x = 2.

We have,

(LHD at x = 2)

 limx→2- f(x) - f(2)x-2 = limh→0 f(2-h) - f(2)-h = limh→0 2h2 -7h + 6 - 6-h =limh→0 -2h + 7 = 7

(RHD at x = 2)

 limx→2+ f(x) - f(2)x-2 = limh→0 f(2+h) - f(2)h = limh→0 10 + 5h - 4 -6h= 5

Thus, LHD at x=2 ≠ RHD at x = 2.

Hence, function is not differentiable at x = 2.

Page No 9.10:

Question 5:

Discuss the continuity and differentiability of the fx=x+x-1 in the interval -1,2

Answer:

Given: fx=x+x-1x=-x for x<0x=x for x>0x-1=-x-1=-x+1 for x-1<0 or x<1x-1=x-1 for x-1>0 or x>1Now,fx=-x-x+1=-2x+1     x∈-1,0orfx=x-x+1=1     x∈0,1orfx=x+x-1=2x-1     x∈1,2
Now,LHL=limx→0- fx=limx→0- -2x+1=0+1=1RHL=limx→0+ fx=limx→0+ 1=1Hence, at x=0, LHL=RHLAgain,LHL=limx→1- fx=limx→1-1=1RHL=limx→1+ fx=limx→1+ 2x-1=2-1=1Hence, at x=1, LHL=RHL
Now,
fx=-x-x+1=-2x+1     x∈-1,0⇒f'x=-2     x∈-1,0orfx=x-x+1=1     x∈0,1⇒f'x=0     x∈0,1orfx=x+x-1=2x-1     x∈1,2⇒f'x=2     x∈1,2
Now,LHL=limx→0- f'x=limx→0- -2=-2RHL=limx→0+ f'x=limx→0+ 0=0Since, at x=0, LHL≠RHLHence, fx is not differentiable at x=0Again,LHL=limx→1- f'x=limx→1- 0=0RHL=limx→1+ f'x=limx→1+ 2=2Since, at x=1, LHL≠RHLHence, fx is not differentiable at x=1

Page No 9.10:

Question 6:

Find whether the function is differentiable at x = 1 and x = 2
fx=xx≤12-x-2+3x-x21≤x≤2x>2

Answer:

fx=xx≤12-x-2+3x-x21≤x≤2x>2⇒f'x=1x≤1-13-2x1≤x≤2x>2
Now,LHL=limx→1- f'x=limx→1- 1=1RHL=limx→1+ f'x=limx→1+ -1=-1Since, at x=1, LHL≠RHLHence, fx is not differentiable at x=1Again,LHL=limx→2- f'x=limx→2- -1=-1RHL=limx→2+ f'x=limx→2+ 3-2x=3-4=-1Since, at x=2, LHL=RHLHence, fx is differentiable at x=2

Page No 9.10:

Question 7:

Show that the function fx=xm sin1x , x≠00, x=0

(i) differentiable at x = 0, if m > 1
(ii) continuous but not differentiable at x = 0, if 0 < m < 1
(iii) neither continuous nor differentiable, if m ≤ 0

Answer:

Given:
        f(x) = xm sin1x0      x≠0 , x=0

(i) Let m=2, then the function becomes f(x) = x2 sin1x0 ,  x≠0, x=0

Differentiability at x=0:
limx→0 f(x) - f(0)x-0 = limx→0 f(x)x = limx→0  x sin1x =0.     
 [ ∵ limx→0 x sin1x = 0 , as x sin1x - 0 = x sin1x = x sin1x ≤x   (∵sinθ≤1  for all θ) and hence x sin1x<0  when x-0<εx-0<ε ]
So, f'(0) = 0, which means f is differentiable at x=0.
Hence the given function is differentiable at x=0.

(ii) Let m=12, 0<m<1. Then the function becomes
     f(x) = x120sin1x  ,     x≠0 , x=0

Continuity at x=0:
(LHL at x=0) = limx→0- f(x) = limh→0 f(0-h) = limh→0 (-h)12 sin10-h = limh→0 h12 sin1h = limh→0 h32  = 0.
(RHL at x=0) = limx→0+ f(x) = limh→0 f(0+h) = limh→0 h12 sin1h = limh→0 h32 = 0.
and f(0) = 0
LHL at x=0 = RHL at x=0 = limx→0 f(x),
Hence continuous.
Now Differentiabilty at x=0 when 0<m<1.
(LHD at x=0) = limx→0- f(x) - f(0)x-0 = limh→0 f(0-h) - f(0)0-h-0 =limh→0 (-h)12 sin1-h-h

 

Page No 9.10:

Question 8:

Find the values of a and b so that the function fxx2+3x+a,if x≤1bx+2       ,if x>1 is differentiable at each x ∈ R.

Answer:

Given:
        f(x) = x2+3x+a,     x≤1bx+2,      x>1 

It is given that the function is differentiable at each x∈R and every differentiable function is continuous.
So, f(x) is continuous at x=1.

Therefore,

   limx→1- f(x)=limx→1+ f(x) = f(1)

⇒limx→1 x2+3x+a = limx→1 bx+2 = a+4                   Using def. of f(x) ⇒ a+4 = b+2 = a+4             ...(i)                                 


Since, f(x) is differentiable at x=1. So,

(LHD at x = 1) = (RHD at x = 1)

    limx→1- f(x) - f(1)x-1 = limx→1+ f(x) - f(1)x-1⇒limx→1 x2+3x+a-a-4x-1 = limx→1 bx+2 -4-ax-1                        Using def. of f(x)⇒ limx→1  (x+4) (x-1)x-1= limx→1 bx-2-ax-1⇒ limx→1  (x+4) (x-1)x-1= limx→1 bx-bx-1                        Using (i)            ⇒ limx→1  (x+4) (x-1)x-1 = limx→1 b(x-1)x-1⇒ 5 = b

From (i), we have

    a+4 = b+2⇒a+4 = 5+2⇒a = 7-4 ⇒a= 3
Hence, a=3 , b=5.

Page No 9.10:

Question 9:

Show that the function fx=2x-3 x,x ≥1sin π x2,x<1 is continuous but not differentiable at x = 1.

Answer:

Given: f(x) = 2x-3 [x] ,  x≥1sinπx2,    x<1

Continuity at x = 1:
(LHL at x = 1) = limx→1-f(x)=limh→0f(1-h)=limh→0 sinπ (1-h)2=sinπ2=1

(RHL at x = 1) = limx→1+ f(x) = limh→0 f(1+h) = limh→0 2(1+h)-31+h=limh→0 2(1+h)-3=1

Hence, (LHL at x = 1) = (RHL at x = 1)

Differentiability at x = 1:

LHD at x=1=limx→1-fx-f1x-1LHD at x=1=limh→0f1-h-f11-h-1LHD at x=1=limh→0f1-h-f1-hLHD at x=1=limh→0sinπ1-h2-1-hLHD at x=1=limh→0cosπh2-1-hLHD at x=1=-π2limh→0cosπh2-1π2h=0RHD at x=1=limx→1+fx-f1x-1RHD at x=1=limh→0f1+h-f11+h-1RHD at x=1=limh→0f1+h-f1hRHD at x=1=limh→0-21+h-3-1hRHD at x=1=limh→0-2hh=-2

LHD ≠ RHD

Hence, the function is continuous but not differentiable at x = 1.



Page No 9.11:

Question 10:

If fx=ax2-b,if x<11x     ,if x≥1 is differentiable at x = 1, find a, b.

Answer:

Given: f(x) =ax2+b,     x<11x,           x≥1
⇒ f(x) = -1x ,                          x<-1ax2-b,                     -1<x<11x,                               x≥1  

It is given that the given function is differentiable at x = 1.

We know every differentiable function is continuous. Therefore it is continuous at x=1. Then,

   lim  x→1- f(x) = limx→1+ f(x) ⇒limx→1 ax2-b = limx→1 1x⇒ a-b = 1                                       ...(i)


It is also differentiable at x=1. Therefore,

 (LHD at x = 1) = (RHD at x = 1)

⇒limx→1-f(x) - f(1)x-1 = limx→1+f(x) - f(1)x-1⇒ limx→1 ax2-b - 1x-1= limx→1 1x - 1x-1 ⇒ limx→1 ax2+1-a-1x-1=limx→1 -(x-1)x-1                 Using (i) ⇒ limx→1 a (x+1)= limx→1 -1 ⇒  2a=-1 ⇒ a =-12
From (i), we have:
          a-b = 1⇒-12 - b = 1⇒ b =- 32

Hence, when a=-12 and b=-32 the function is differentiable at x = 1. 

Page No 9.11:

Question 11:

Find the values of a and b, if the function f defined by fx=x2+3x+a,x⩽1bx+2,x>1 is differentiable at x = 1.

Answer:

Given that f(x) is differentiable at x = 1. Therefore,  f(x) is continuous at x = 1.

limx→1-fx=limx→1+fx=f1⇒limx→1x2+3x+a=limx→1bx+2=1+3+a⇒1+3+a=b+2⇒a-b+2=0                              .....1

Again, f(x) is differentiable at x = 1. So,

(LHD at x = 1) = (RHD at x = 1)
⇒limx→1-fx-f1x-1=limx→1+fx-f1x-1
limx→1x2+3x+a-4+ax-1=limx→1bx+2-4+ax-1⇒limx→1x2+3x-4x-1=limx→1bx-2-ax-1⇒limx→1x+4x-1x-1=limx→1bx-bx-1⇒limx→1x+4=limx→1bx-1x-1⇒5=b

Putting b = 5 in (1), we get

a = 3

Hence, a = 3 and b = 5.



Page No 9.16:

Question 1:

If f is defined by f (x) = x2, find f'(2).

Answer:

Given: f(x) = x2.

We know a  polynomial function is everywhere differentiable. Therefore f(x) is differentiable at x=2.

 f'(2) = limh→0 f(2+h) - f(2)h⇒ f'(2) = limh→0 (2+h)2 - 22h⇒ f'(2) = limh→0 (4+h2+4h) - 4h⇒ f'(2) = limh→0 h (h+4)h⇒ f'(2) = 4

Page No 9.16:

Question 2:

If f is defined by fx=x2-4x+7, show that f'5=2f'72

Answer:

Given: f(x) = x2-4x+7

Clearly, f(x) being a polynomial function, is everywhere differentiable. The derivative of f at x is given by:

 f'(x) = limh→0 f(x+h) - f(x)h⇒ f'(x) = limh→0 x+h2 -4(x+h) +7 - (x2 -4x+7)h⇒ f'(x) = limh→0 x2+h2+2xh -4x-4h+7 -x2+4x-7h⇒ f'(x) = limh→0 h2 +2xh -4hh⇒ f'(x) = limh→0 h(h+2x-4)h⇒ f'(x) = 2x-4

Now, 

f'(5) = 2×5 - 4= 6f'72 = 2×72 - 4 = 3
Therefore,   f'(5) = 2×3 = 2f'72
 Hence proved.

Page No 9.16:

Question 3:

Show that the derivative of the function f given by
fx=2x3-9x2+12x+9, at x = 1 and x = 2 are equal.

Answer:

Given: f(x) = 2x3-9x2+12x+9

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of f at x is given by:

   f'(x) = limh→0 f(x+h - f(x)h⇒ f'(x) = limh→0  2(x+h)3-9(x+h)2+12(x+h) + 9 - 2x3+9x2-12x-9h⇒ f'(x) = limh→0 2x3 + 2h3+6x2h +6xh2 -9x2-9h2-18xh+12x+12h+9 -2x3+9x2-12x-9h⇒ f'(x) = limh→0 2h3 +6x2h +6xh2 -9h2 -18xh+12hh⇒ f'(x) = limh→0  h(h2 +6x2+6xh -9h-18x+12)h⇒ f'(x) = 6x2-18x+12

So,

f'(1) = 6x2-3x+2          = 6×(1-3+2)          = 0f'(2) = 6x2-3x+2          = 6×(4-6+2)          = 0

Hence the derivative at x=1 and x=2 are equal.

Page No 9.16:

Question 4:

If for the function Φ x=λx2+7x-4, Φ'5=97, find λ.

Answer:

Given: ϕ(x) = λx2+7x-4

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of ϕ at x is given by:

ϕ'(x) = limh→0 ϕ(x+h) - ϕ(x)h⇒ ϕ'(x) =limh→0 λ(x+h)2 +7(x+h) -4 - λx2-7x+4h⇒ ϕ'(x) = limh→0 λx2 +λh2+2λxh+7x+7h-4-λx2-7x+4h⇒ ϕ'(x) = limh→0 λh2 +2λxh+7hh⇒ ϕ'(x) = limh→0 h(λh +2λx+7)h⇒ ϕ'(x) = 2λx+7
It is given ϕ'(5) = 97
Thus,
      ϕ'(5) = 10λ + 7 = 97⇒ 10λ+7 =97⇒ 10λ = 90⇒ λ = 9

Page No 9.16:

Question 5:

If fx=x3+7x2+8x-9, find f'(4).

Answer:

Given: f(x) = x3+7x2+8x-9

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of f at x is given by:

 f'(x) = limh→0f(x+h) - f(x)h⇒ f'(x) = limh→0 (x+h)3+7(x+h)2+8(x+h)-9 - x3 -7x2-8x+9h⇒ f'(x) = limh→0 x3+h3+3x2h + 3xh2+7x2+7h2+14xh+8x+8h-9-x3-7x2-8x+9h⇒ f'(x) = limh→0 h3+3x2h+3xh2+7h2+14xh+8hh⇒ f'(x) = limh→0 h(h2+3x2+3xh+7h+14x+8)h⇒ f'(x) = limh→0 h2+3x2+3xh+7h+14x+8⇒ f'(x) = 3x2+14x+8
Thus,
      f'(4) = 3×42+14×4+8          = 48+56+8         =112

Page No 9.16:

Question 6:

Find the derivative of the function f defined by f (x) = mx + c at x = 0.

Answer:

Given: f(x) = mx+c

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of f at x is given by:

f'(x) = limh→0 f(x+h) - f(x)h⇒  f'(x) = limh→0 m(x+h) +c -mx-ch⇒ f'(x) = limh→0 mx+mh+c-mx-ch⇒ f'(x) = limh→0 mhh ⇒ f'(x) = m

Thus, f'(0) = m

Page No 9.16:

Question 7:

Examine the differentialibilty of the function f defined by
fx=2x+3if-3≤ x≤-2x+1x+2if -2≤x<0if 0≤x≤1

Answer:

fx=2x+3if-3≤ x≤-2x+1x+2if -2≤x<0if 0≤x≤1⇒f'x=2if-3≤ x≤-211if -2≤x<0if 0≤x≤1
Now,LHL=limx→-2- f'x=limx→-2- 2=2RHL=limx→-2+ f'x=limx→-2+ 1=1Since, at x=-2, LHL≠RHLHence, fx is not differentiable at x=-2Again,LHL=limx→0- f'x=limx→0- 1=1RHL=limx→0+ f'x=limx→0+ 1=1Since, at x=0, LHL=RHLHence, fx is differentiable at x=0

Page No 9.16:

Question 8:

Write an example of a function which is everywhere continuous but fails to differentiable exactly at five points.

Answer:

fx=x+x+1+x+2+x+3+x+4
The above function is continuous everywhere but not differentiable at x = 0, 1, 2, 3 and 4

Page No 9.16:

Question 9:

Discuss the continuity and differentiability of f (x) = |log |x||.

Answer:

We have,
f (x) = |log |x||

x=-x        -∞<x<-1-x        -1<x<0  x        0<x<1  x        1<x<∞log x=log -x        -∞<x<-1log -x        -1<x<0  log x        0<x<1 log x        1<x<∞log x=log -x        -∞<x<-1-log -x        -1<x<0 -log x        0<x<1 log x        1<x<∞
LHD at x=-1=limx→-1-fx-f-1x+1                             =limx→-1-log -x-0x+1                             =limh→0log 1+h-1-h+1                             =-limh→0log 1+hh=-1

RHD at x=-1=limx→-1+fx-f-1x+1                             =limx→-1+-log -x-0x+1                             =limh→0-log 1-h-1+h+1                             =limh→0-log 1-hh=1
Here, LHD ≠ RHD
So, function is not differentiable at x = − 1

At 0 function is not defined.
LHD at x=1=limx→1-fx-f1x-1                        =limx→1--log x-0x-1                       =limh→0-log 1-h1-h-1                       =-limh→0log 1-hh=-1
RHD at x=1=limx→1+fx-f1x-1                        =limx→1+log x-0x-1                       =limh→0log 1+h1+h-1                       =limh→0log 1+hh=1
Here, LHD ≠ RHD
So, function is not differentiable at x = 1
Hence, function is not differentiable at x = 0 and ± 1
At 0 function is not defined.
So, at 0 function is not continuous.
LHL at x=-1=limx→-1-fx                             =limx→-1-log -x                             =log 1=0RHL at x=-1=limx→-1+fx                             =limx→-1+-log -x                             =-log 1=0f-1=0Therefore, fx=log x is continuous at x=-1
LHL at x=1=limx→1-fx                        =limx→1--log x                        =-log 1=0RHL at x=1=limx→1+fx                        =limx→1+log x                        =log 1=0f1=0Therefore, at x=1, fx=log x is continuous.

Hence, function f (x) = |log |x|| is not continuous at x = 0

Page No 9.16:

Question 10:

Discuss the continuity and differentiability of f (x) = e|x| .

Answer:

Given: f(x) = ex
      ⇒ f(x) = ex,           x≥0e-x,         x<0

Continuity: 

(LHL at x = 0) 

limx→0- f(x) = limh→0 f(0-h) = limh→0 e-(0-h)  = limh→0 eh = 1

(RHL at x = 0) 

limx→0+ f(x) = limh→0 f(0+h) = limh→0 e(0+h) = 1

and f(0) = e0 = 1

Thus,  limx→0- f(x) = limx→0+ f(x) = f(0)

Hence,function is continuous at x = 0 .

Differentiability at x = 0.

(LHD at x = 0)

 limx→0-f(x) - f(0)x-0= limh→0f(0-h) - f(0)0-h-0= limh→0 e-(0-h) - 1-h= limh→0 eh-1-h  = -1                   ∵ limh→0 eh-1h =1 
(RHD at x = 0)

 limx→0+f(x) - f(0)x-0= limh→0f(0+h) - f(0)0+h-0= limh→0 e(0+h) - 1h= limh→0 eh-1h  = 1                   ∵ limh→0 eh-1h =1 

LHD at (x = 0)≠RHD at (x = 0)

Hence the function is not differentiable at x = 0.

Page No 9.16:

Question 11:

Discuss the continuity and differentiability of

fx=x-c cos 1x-c,x≠c0                                ,x=c

Answer:

Given: fx=x-c cos 1x-c,x≠c0                                ,x=c

Continuity:

(LHL at x = c) 

limx→c- f(x) = limh→0 f(c-h) = limh→0 (c-h-c)  cos1c-h-c = limh→0 -h cos1h Since , cos 1h  is a bounded function and 0 ×bounded function is 0=0

(RHL at x = c) 

limx→c+ f(x) =limh→0 f(c+h) = limh→0 (c+h-c) cos1c+h-c = limh→0 h cos1h Since, cos1h  is a bounded function and 0× bounded function is 0=0

and 
Differentiability at x = c

(LHD at x = c)
limx→c- f(x) - f(c)x-c = limh→0f(c-h) - f(c)c-h-c = limh→0 -h cos1-h -  0-h           ∵0.cos 1c-c=0, as cos function is bounded function.=lim h→0cos1h=A number which oscillates between -1 and 1∴LHD (x=c) does not exist . Similarly , we can show that RHD(x=c) does not exist . Hence , f(x) is not differentiable at x=c
 

Page No 9.16:

Question 12:

Is |sin x| differentiable? What about cos |x|?

Answer:

Let, f(x) = |sin x|

sin x=0,  for x=nπ,sin x=-sin x      2m-1π< x<2mπ , where m∈Z sin x       2mπ< x<2m+1π , where m∈Z- sin x       2m+1π< x<2m+1π , where m∈Z

LHD at x=2mπ=limx→2mπ-fx-f2mπx-2mπ                             =limx→2mπ--sinx-0x-2mπ                             =limh→0-sin2mπ-h2mπ-h-2mπ                             =limh→0sinh-h=-1
RHD at x=2mπ=limx→2mπ+fx-f2mπx-2mπ                             =limx→2mπ+sinx-0x-2mπ                             =limh→0sin2mπ+h2mπ+h-2mπ                             =limh→0sinhh=1


Here, LHD ≠ RHD So, function is not differentiable at x=2mπ, where, m∈Z     .....1                                                                                          
LHD at x=2m+1π=limx→2m+1π-fx-f2m+1πx-2m+1π                                   =limx→2m+1π-sin x-0x-2m+1π                                   =limh→0sin 2m+1π-h2m+1π-h-2m+1π                                   =limh→0sin h-h=-1
RHD at x=2m+1π=limx→2m+1π+fx-f2m+1πx-2m+1π                                   =limx→2m+1π+-sin x-0x-2m+1π                                   =limh→0-sin 2m+1π+h2m+1π+h-2m+1π                                   =limh→0sin hh=1
Here, LHD ≠ RHD. So, function is not differentiable at x=2m+1π, where, m∈Z     .....2                                                                                         
From, 1 and 2, we getfx=sin x is not differentiable at x=nπ

We know that,cos x=cos x               For all x∈RAlso we know that cos x is differentiable at all real points.Therefore, cos x is differentiable everywhere.



Page No 9.17:

Question 5:

Let fx=x+x x. Then, for all x
(a) f is continuous
(b) f is differentiable for some x
(c) f' is continuous
(d) f'' is continuous

Answer:

(a) f is continuous
(c) f' is continuous

We have,fx=x+x x       =xx+x2       =xx+x2fx=2x2          x≥00             x<0To check continuity of fx at x=0LHL at x=0=limx→0-fx                       =limx→0-0                       =0RHL at x=0=limx→0+fx                       =limx→0+2x2                       =0And f0=0Here, LHL=RHL=f0Therefore, fx is continuous at x=0Hence, fx is continuous everywhere.

To check the differentiability of fx at x=0LHD at x=0=limx→0-fx-f0x-0                        =limx→0-0-0x=0RHD at x=0=limx→0+fx-f0x-0                        =limx→0-2x2-0x                        =limx→0-2x2-0x                        =limx→0-2x=0LHD=RHDTherefore, fx is derivative at x=0Hence, fx is differentiable everywhere.

f'x=4x         x≥00             x<0To check continuity of f'x at x=0LHL at x=0=limx→0-f'x                       =limx→0-0                       =0RHL at x=0=limx→0+f'x                       =limx→0+4x                       =0And f'0=0Here, LHL=RHL=f'0Therefore, f'x is continuous at x=0Hence, f'x is continuous everywhere.
f''x=4         x≥00             x<0To check continuity of f''x at x=0LHL at x=0=limx→0-f''x                       =limx→0-0                       =0RHL at x=0=limx→0+f''x                       =limx→0+4                       =4Therefore, LHL≠RHL Therefore, f''x is not continuous at x=0Hence, f''x is not continuous everywhere.

Page No 9.17:

Question 6:

The function f (x) = e−|x| is
(a) continuous everywhere but not differentiable at x = 0
(b) continuous and differentiable everywhere
(c) not continuous at x = 0
(d) none of these

Answer:

(a) continuous everywhere but not differentiable at x = 0



Given: f(x) = e-x = ex,                  x ≥0e-x,                x<0Continuity :limx→0- f(x) = limh→0 f(0-h) = limh→0 e-(0-h)  = limh→0 eh = 1  

RHL at x = 0

limx→0+ f(x) = limh→0 f(0+h) = limh→0 e(0+h) = 1

and f(0) = f(0) = e0 = 1
Thus, limx→0- f(x) = limx→0+ f(x) = f

Hence, function is continuous at x = 0

Differentiability at x = 0

(LHD at x = 0)

limx→0-f(x) - f(0)x-0= limh→0f(0-h) - f(0)0-h-0= limh→0 e-(0-h) - 1-h= limh→0 ehh = ∞

Therefore, left hand derivative does not exist.
Hence, the function is not differentiable at x = 0.

Page No 9.17:

Question 7:

The function f (x) = |cos x| is
(a) everywhere continuous and differentiable
(b) everywhere continuous but not differentiable at (2n + 1) π/2, n ∈ Z
(c) neither continuous nor differentiable at (2n + 1) π/2, n ∈ Z
(d) none of these

Answer:

(b) everywhere continuous but not differentiable at (2n + 1) π/2, n ∈ Z

We have,fx=cos x⇒fx=cos x,      2nπ≤x<4n+1π20,     x=4n+1π2    -cos x,    4n+1π2<x<4n+3π2             0 ,        x=4n+3π2        cos x,        4n+3π2< x≤2n+2πWhen, x is in first quadrant, i.e. 2nπ≤x<4n+1π2 , we have         fx=cos x which being a trigonometrical function is continuous and differentiable in 2nπ, 4n+1π2When, x is in second quadrant or in third quadrant, i.e., 4n+1π2<x<4n+3π2 , we have         fx=-cos x which being a trigonometrical function is continuous and differentiable in 4n+1π2, 4n+3π2When, x is in fourth quadrant, i.e., 4n+3π2< x≤2n+2π , we have         fx=cos x which being a trigonometrical function is continuous and differentiable in 4n+3π2, 2n+2πThus possible point of non-differentiability of fx are x=4n+1π2, 4n+3π2Now, LHD at x=4n+1π2 =limx→4n+1π2- fx- f4n+1π2x-4n+1π2                                               =limx→4n+1π2-  cos x- 0x-4n+1π2                                              =limx→4n+1π2-  -sin x1-0      By L'Hospital rule                                              =-1And  RHD at x=4n+1π2 =limx→4n+1π2+ fx- f4n+1π2x-4n+1π2                                              =limx→4n+1π2+  -cos x- 0x-4n+1π2                                              =limx→4n+1π2+  sin x1-0      By L'Hospital rule                                              =1∴lim x→4n+1π2-fx ≠limx→4n+1π2+fxSo fx is not differentiable at x=4n+1π2Now, LHD at x=4n+3π2 =limx→4n+1π2- fx- f4n+3π2x-4n+3π2                                              =limx→4n+3π2-  -cos x- 0x-4n+3π2                                              =limx→4n+3π2-  sin x1-0      By L'Hospital rule                                              =1And  RHD at x=4n+3π2 =limx→4n+3π2+ fx- f4n+3π2x-4n+3π2                                              =limx→4n+3π2+  cos x- 0x-4n+3π2                                              =limx→4n+3π2+ -sin x1-0      By L'Hospital rule                                              =-1∴lim x→4n+3π2-fx ≠limx→4n+3π2+fxSo fx is not differentiable at x=4n+3π2Therefore, fx is neither differentiable at 4n+1π2 nor at 4n+3π2i.e. fx is not differentiable at odd multiples of π2i.e. fx is not differentiable at x=2n+1π2Therefore, f(x) is everywhere continuous but not differentiable at 2n+1π2 .

Page No 9.17:

Question 8:

If fx=1-1-x2, then f x is

(a) continuous on [−1, 1] and differentiable on (−1, 1)
(b) continuous on [−1, 1] and differentiable on -1, 0∪0, 1
(c) continuous and differentiable on [−1, 1]
(d) none of these

Answer:

b continuous on -1,1 and differentiable on -1,0∪0,1

We have,fx=1-1-x2Here, function will be defined for those values of x for which1-x2≥0⇒1≥x2⇒x2≤1⇒x≤1⇒-1≤x≤1Therefore, function is continuous in -1, 1
Now, we need to check the differentiability of fx=1-1-x2 in the interval -1, 1.Now, we will check the differentiability at x=0LHD at x=0=limx→0-fx-f0x-0                        =limx→0-1-1-x2-0x                        =limx→0-1-1-x2x                        =limh→01-1-0-h20-h                        =limh→01-1-h2-h=-∞RHD at x=0=limx→0+fx-f0x-0                        =limx→0+1-1-x2-0x                        =limx→0+1-1-x2x                        =limh→01-1-0+h20+h                        =limh→01-1-h2h=∞ So, the function is not differentiable at x=0.                

Page No 9.17:

Question 9:

If fx=asin x+bex+cx3 and if f (x) is differentiable at x = 0, then
(a) a=b=c=0
(b) a=0, b=0; c∈R
(c) b=c=0, a∈R
(d) c=0, a=0, b∈R

Answer:

(b) a=0, b=0; c∈R

We have,fx=a sin x+be x+c x3       =a sin x+bex+cx3           0<x<π2-a sin x+be-x-cx3     -π2<x<0Here, fx is differentiable at x=0Therefore, LHD at x=0=RHD at x=0⇒limx→0-fx-f0x-0=limx→0+fx-f0x-0⇒limx→0--a sinx+be-x-cx3-bx=limx→0+a sin x+bex+cx3-bx⇒limh→0-a sin0-h+be-0-h-c0-h3-b0-h=limh→0a sin 0+h+be0+h+c0+h3-b0+h⇒limh→0a sin h+be h+ch3-b-h=limh→0a sin h+beh+ch3-bh⇒limh→0a cos h+beh+3ch2-1=limh→0a cos h+beh+3ch21         By L'Hospital rule⇒-a+b=a+b ⇒-2a+b=0⇒a+b=0This is true for all value of c∴c∈RIn the given options, option b satisfies a+b=0 and c∈R

Page No 9.17:

Question 1:

Let f (x) = |x| and g (x) = |x3|, then
(a) f (x) and g (x) both are continuous at x = 0
(b) f (x) and g (x) both are differentiable at x = 0
(c) f (x) is differentiable but g (x) is not differentiable at x = 0
(d) f (x) and g (x) both are not differentiable at x = 0

Answer:

Option (a) f (x) and g (x) both are continuous at x = 0

Given: fx =  x ,  gx =  x3 

We know  x is continuous at x=0 but not differentiable at x = 0 as (LHD at x = 0) ≠ (RHD at x = 0).
Now, for the function gx =  x3  = x3,             x≥0-x3,          x<0
Continuity at x = 0:

(LHL at x = 0) = limx→0-gx = limh→0 g0-h = limh→0--h3 = limh→0 h3 = 0.

(RHL at x = 0) = limx→0+fx = limh→0f0+h = limh→0 h3 = 0.

and g0 = 0.
Thus,   limx→0-gx = limx→0+gx = g0.
Hence, g(x) is continuous at x = 0.

Differentiability at x = 0:

(LHD at x = 0) = limx→0-fx - f0x-0 = limh→0f0-h - f00-h-0 = limh→0h3 -0-h = 0.

(RHD at x = 0) = limx→c+fx - f0x-0 = limh→0f0+h - f00+h-0 = limh→0h3 -0h = limh→0h3h =0
Thus, (LHD at x = 0) = (RHD at x = 0). 
Hence, the function  gx is differentiable at x = 0.

Page No 9.17:

Question 2:

The function f (x) = sin−1 (cos x) is
(a) discontinuous at x = 0
(b) continuous at x = 0
(c) differentiable at x = 0
(d) none of these

Answer:

(b) continuous at x = 0

Given: f(x) = sin-1cos x.

Continuity at x = 0:

We have,
(LHL at x = 0) 

limx→0- f(x) =limh→0 sin-1cos0-h = limh→0 sin-1cos h= sin-11= π2

(RHL at x = 0)

  limx→0+ fx= limh→0 sin-1cos0+h= limh→0 sin-1cos h = sin-11 = π2

f(0) = sin-1cos 0        = sin-11       = π2

Differentiability at x = 0:
(LHD at x = 0) 

limx→0- fx - f0x-0 = limh→0 sin-1cos0-h - π2-h = limh→0 sin-1cos-h -π2-h = limh→0 sin-1cosh -π2-h= limh→0 sin-1sin π2-h -π2-h= limh→0-h-h=1

RHD at x = 0

limx→0+ fx - f0x-0 = limh→0 sin-1cos0+h-π2h = limh→0 sin-1cosh-π2h = limh→0 sin-1sin π2-h-π2-h= limh→0-hh=-1

∴ LHD≠RHD

Hence, the function is not differentiable at x = 0 but is continuous at x = 0.

Page No 9.17:

Question 3:

The set of points where the function f (x) = x |x| is differentiable is
(a) -∞, ∞
(b) -∞, 0∪0, ∞
(c) 0, ∞
(d) 0, ∞

Answer:

(a) -∞, ∞

We have,fx=xx⇒fx=-x2,        x<0   0 ,        x=0   x2,         x>0When, x<0, we have         fx=-x2 which being a polynomial function is continuous and differentiable in -∞, 0When, x>0, we have         fx=x2 which being a polynomial function is continuous and differentiable in 0, ∞Thus possible point of non-differentiability of fx is x=0Now, LHD at x=0 =limx→0- fx- f0x-0                              =limx→0--x2- 0x                              =limh→0--h2-h                              =limh→0h                              =0And  RHD at x=0 =limx→0+ fx- f0x-0                              =limx→0+x2- 0x                              =limh→0h2h                              =lim h→0h                              =0∴LHD at x=0=RHD at x=0So, fx is also differentiable at x=0i.e. fx is differentiable in -∞, ∞

Page No 9.17:

Question 4:

If fx=x+2tan-1 x+2, x≠-22, x=-2, then f (x) is

(a) continuous at x = − 2
(b) not continuous at x = − 2
(c) differentiable at x = − 2
(d) continuous but not derivable at x = − 2

Answer:

(b) not continuous at x = − 2

Given:
 f(x) = x+2tan-1(x+2) ,    x≠-2                  2  ,               x=-2
⇒ f(x) = -(x+2)tan-1(x+2),         x<-2 (x+2)tan-1(x+2),      x>-22  ,                      x=-2
Continuity at x = − 2.
(LHL at x= − 2) = limx→-2-f(x)=limh→0f(-2-h)=limh→0-(-2-h+2)tan-1(-2-h+2)=limh→0htan-1(-h) =-1.

(RHL at x = −2) = limx→-2+f(x=limh→0f(-2+h=limh→0(-2+h+2)tan-1(-2+h+2)=limh→0htan-1(h) =1.

Also f(-2) = 2
Thus,  limx→-2-f(x)≠limx→-2+ f(x) ≠ f(-2).
Therefore, given function is not continuous at x = − 2



Page No 9.18:

Question 10:

If fx=x2+x21+x2+x21+x2+...+x21+x2+....,

then at x = 0, f (x)
(a) has no limit
(b) is discontinuous
(c) is continuous but not differentiable
(d) is differentiable

Answer:

(b) is discontinuous

We have,fx=x2+x21+x2+x21+x2+...+x21+x2+....,When x=0 then x2=0 and x21+x2=0∴f0=0+0+0+0.......⇒f0=0When, x≠0Then, x2>0and 1+x2>x2⇒0<x21+x2<1∴limx→0 fx=lim x→0x2+x21+x2+x21+x2+...+x21+x2+....,              =limx→0x21+11+x2+11+x2+...+11+x2+....,              =limx→0x211-11+x2             Sum of infinite series where, r=11+x2              =limx→0x21+x2x2              =limx→01+x2              =1∴limx→0 fx≠f0∴fx is discontinuous at x=0

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Question 11:

If fx=logex, then
(a) f' 1+=1
(b) f' 1=-1
(c) f' 1=1
(d) f' 1=-1

Answer:

(a) f' 1+=1 and (b) f' 1=-1

fx=logex, = - loge x , for 0 < x < 1loge x , for x ≥1Differentiablity at x = 1,we have ,   (LHD at x = 1 ) =lim x →1- f(x) - f(1)x -1                           = lim x →1- -log x - log 1x -1                             = -lim x →1- log x x -1                                                    = -lim h →0 log (1 - h) 1 -h - 1                            = -lim h →0 log (1 - h)  -h   = -1                                      

    (RHD at x = 1 )  = lim x →1+ f(x) - f(1)x -1                                                                 = lim x →1+ log x - log (1)x -1                                                               = lim h → 0  log (1 + h)x -1 =  lim h → 0  log (1 + h)h = 1

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Question 12:

If fx=loge |x|, then
(a) f (x) is continuous and differentiable for all x in its domain
(b) f (x) is continuous for all for all × in its domain but not differentiable at x = ± 1
(c) f (x) is neither continuous nor differentiable at x = ± 1
(d) none of these

Answer:

(b) f (x) is continuous for all x in its domain but not differentiable at x = ± 1

We have,fx=loge |x|We know that log function is defined for positive value.Here, x is positive for all non zero x.Therefore, domain of function is R-0
And we know that logarithmic function is continuous in its domain.
Therefore, logex is continuous in its domain.
We will check the differentiability at its critical points. logex=loge-x                     -∞<x<-1-loge-x                    -1<x<0-logex                       0<x<1logex                    1<x<∞LHD at x=-1=limx→-1-fx-f-1x--1                         =limx→-1-loge-x-0x+1                         =limh→0loge--1-h-1-h+1                         =limh→0loge1+h-h                         =-1RHD at x=-1=limx→-1+fx-f-1x--1                         =limx→-1+-loge-x-0x+1                          =limh→0-loge--1+h-1+h+1                          =limh→0-loge1-hh                            =-limh→0loge1-hh                            =-1×-1=1Here, LHD≠RHDTherefore, the given function is not differentiable at x=-1.

LHD at x=1=limx→1-fx-f1x-1                      =limx→1--logex-0x-1                      =limh→0-loge1-h1-h-1                      =limh→0loge1-hh                      =-1RHD at x=1=limx→1+fx-f1x-1                      =limx→1+logex-0x-1                      =limh→0loge1+h1+h-1                      =limh→0loge1+hh                      =1Here, LHD≠RHDTherefore,the given function is not differentiable at x=1.
Therefore, given function is continuous for all x in its domain but not differentiable at x = ± 1

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Question 13:

Let fx=1xfor x≥1ax2+bfor x<1

If f (x) is continuous and differentiable at any point, then
(a) a=12, b=-32
(b) a=-12, b=32
(c) a = 1, b = − 1
(d) none of these

Answer:

(b) a=-12, b=32

We have,fx=-1x,       x≤-1ax2+b,    -1<x<11x,          x≥1Given: fx is differentiable and continuous at every point.Consider a point x=1limx→1-fx=limx→1+fx⇒limx→1-ax2+b=limx→1+1x⇒a+b=1          ...iIt is also differentiable at x=1limx→1-fx-f1x-1=limx→1+fx-f1x-1⇒limx→1-ax2+b-1x-1=limx→1+1x-1x-1⇒limx→1-ax2-ax-1=limx→1+1-xx-1x       Using i⇒limx→1-ax+1=limx→1+-x⇒2a=-1⇒a=-12Plugging a=-12 in i we get,b=32∴a=-12, b=32

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Question 14:

The function f (x) = x − [x], where [⋅] denotes the greatest integer function is
(a) continuous everywhere
(b) continuous at integer points only
(c) continuous at non-integer points only
(d) differentiable everywhere

Answer:

(c) continuous at non-integer points only

We have,fx=x-xConsider n be an integer.fx=x-x=x-n-1       n-1≤x<n0                   x=nx-n             n≤x<n+1Now,LHL at x=n=limx→n-fx=x-n-1=x-n+1RHL at x=n=limx→n+fx=x-n=x-nAs, LHL≠RHL at x=ni.e., given function is not continuous at n.Now, n is any integer.Therefore, given function is not continuous at integers.

Therefore, given points are continuous at non-integer points only.

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Question 15:

Let fxax2+1,x>1x+1/2,x≤1. Then, f (x) is derivable at x = 1, if

(a) a = 2
(b) a = 1
(c) a = 0
(d) a = 1/2

Answer:

(d) a = 1/2


Given: f(x) = ax2+1,   x>1x+12,     x≤1
The function is derivable at x = 1, iff left hand derivative and right hand derivative of the function are equal at x = 1.

LHD at x=1=limx→1-fx-f1x-1LHD at x=1=limh→0f1-h-f11-h-1LHD at x=1=limh→0f1-h-f1-hLHD at x=1=limh→01-h+12-32-h=1RHD at x=1=limx→1+fx-f1x-1RHD at x=1=limh→0f1+h-f11+h-1RHD at x=1=limh→0f1+h-f1hRHD at x=1=limh→0a1+h2+1-32hRHD at x=1=limh→0a1+h2+2h-12h∵ LHD=RHD⇒a-12=0⇒a=12

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Question 16:

Let f (x) = |sin x|. Then,
(a) f (x) is everywhere differentiable.
(b) f (x) is everywhere continuous but not differentiable at x = n π, n ∈ Z
(c) f (x) is everywhere continuous but not differentiable at x=2n+1π2, n∈Z.
(d) none of these

Answer:

(b) f (x) is everywhere continuous but not differentiable at x = n π, n ∈ Z

We have,fx=sin x⇒fx=0 ,           x=2nπ sin x,      2nπ<x<2n+1π0,            x=2n+1π-sin x,   2n+1π<x<2n+2πWhen, x is in first or second quadrant, i.e., 2nπ<x<2n+1π , we have         fx=sin x which being a trigonometrical function is continuous and differentiable in 2nπ, 2n+1πWhen, x is in third or fourth quadrant, i.e., 2n+1π<x<2n+2π , we have         fx=-sin x which being a trigonometrical function is continuous and differentiable in 2n+1π, 2n+2πThus possible point of non-differentiability of fx are x=2nπ and  2n+1πNow, LHD at x=2nπ =limx→2nπ- fx- f2nπx-2nπ                                     =limx→2nπ-  -sin x- 0x-2nπ                                     =limx→2nπ-  -cos x1-0      By L'Hospital rule                                     =-1And  RHD at x=2nπ =limx→2nπ+ fx- f2nπx-2nπ                                   =limx→2nπ+  sin x- 0x-2nπ                                   =limx→2nπ+  cos x1-0      By L'Hospital rule                                   =1∴lim x→2nπ-fx ≠limx→2nπ+fxSo fx is not differentiable at x=2nπNow, LHD at x=2n+1π =limx→2n+1π- fx- f2n+1πx-2n+1π                                            =limx→2n+1π-  sin x- 0x-2n+1π                                            =limx→2n+1π-  cos x1-0      By L'Hospital rule                                           =-1And  RHD at x=2n+1π =limx→2n+1π+ fx- f2n+1πx-2n+1π                                           =limx→2n+1π+  -sin x- 0x-2n+1π                                           =limx→2n+1π+ -cos x1-0      By L'Hospital rule                                            =1∴lim x→2n+1π-fx ≠limx→2n+1π+fxSo fx is not differentiable at x=2n+1πTherefore, fx is neither differentiable at 2nπ nor at 2n+1πi.e.fx is neither differentiable at even multiple of π nor at odd multiple of πi.e. fx is not differentiable at x=nπTherefore, f(x) is everywhere continuous but not differentiable at nπ.

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Question 17:

Let f (x) = |cos x|. Then,
(a) f (x) is everywhere differentiable
(b) f (x) is everywhere continuous but not differentiable at x = n π, n ∈ Z
(c) f (x) is everywhere continuous but not differentiable at x=2n+1π2, n∈Z.
(d) none of these

Answer:

(c) f (x) is everywhere continuous but not differentiable at x=2n+1π2, n∈Z.

We have,fx=cos x⇒fx=cos x,      2nπ≤x<4n+1π20,     x=4n+1π2    -cos x,    4n+1π2<x<4n+3π2             0 ,        x=4n+3π2        cos x,        4n+3π2< x≤2n+2πWhen, x is in first quadrant, i.e. 2nπ≤x<4n+1π2 , we have         fx=cos x which being a trigonometrical function is continuous and differentiable in 2nπ, 4n+1π2When, x is in second quadrant or in third quadrant, i.e., 4n+1π2<x<4n+3π2 , we have         fx=-cos x which being a trigonometrical function is continuous and differentiable in 4n+1π2, 4n+3π2When, x is in fourth quadrant, i.e., 4n+3π2< x≤2n+2π , we have         fx=cos x which being a trigonometrical function is continuous and differentiable in 4n+3π2, 2n+2πThus possible point of non-differentiability of fx are x=4n+1π2, 4n+3π2Now, LHD at x=4n+1π2 =limx→4n+1π2- fx- f4n+1π2x-4n+1π2                                               =limx→4n+1π2-  cos x- 0x-4n+1π2                                              =limx→4n+1π2-  -sin x1-0      By L'Hospital rule                                              =-1And  RHD at x=4n+1π2 =limx→4n+1π2+ fx- f4n+1π2x-4n+1π2                                              =limx→4n+1π2+  -cos x- 0x-4n+1π2                                              =limx→4n+1π2+  sin x1-0      By L'Hospital rule                                              =1∴lim x→4n+1π2-fx ≠limx→4n+1π2+fxSo fx is not differentiable at x=4n+1π2Now, LHD at x=4n+3π2 =limx→4n+1π2- fx- f4n+3π2x-4n+3π2                                              =limx→4n+3π2-  -cos x- 0x-4n+3π2                                              =limx→4n+3π2-  sin x1-0      By L'Hospital rule                                              =1And  RHD at x=4n+3π2 =limx→4n+3π2+ fx- f4n+3π2x-4n+3π2                                              =limx→4n+3π2+  cos x- 0x-4n+3π2                                              =limx→4n+3π2+ -sin x1-0      By L'Hospital rule                                              =-1∴lim x→4n+3π2-fx ≠limx→4n+3π2+fxSo fx is not differentiable at x=4n+3π2Therefore, fx is neither differentiable at 4n+1π2 nor at 4n+3π2i.e. fx is not differentiable at odd multiples of π2i.e. fx is not differentiable at x=2n+1π2Therefore, f(x) is everywhere continuous but not differentiable at 2n+1π2 .
 

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Question 18:

The function f (x) = 1 + |cos x| is
(a) continuous no where
(b) continuous everywhere
(c) not differentiable at x = 0
(d) not differentiable at x = n π, n ∈ Z

Answer:

(b) continuous everywhere

Graph of the function f (x) = 1 + |cos x| is as shown below:



From the graph, we can see that f (x) is everywhere continuous but not differentiable at x=2n+1π2, n∈Z

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Question 19:

The function f (x) =  |cos x| is
(a) differentiable at x = (2n + 1) π/2, n ∈ Z
(b) continuous but not differentiable at x = (2n + 1) π/2, n ∈ Z
(c) neither differentiable nor continuous at x = n ∈ Z
(d) none of these

Answer:

(b) continuous but not differentiable at x = (2n + 1) π/2, n ∈ Z

We have,fx=cos x⇒fx=cos x,      2nπ≤x<4n+1π20,     x=4n+1π2    -cos x,    4n+1π2<x<4n+3π2             0 ,        x=4n+3π2        cos x,        4n+3π2< x≤2n+2πWhen, x is in first quadrant, i.e. 2nπ≤x<4n+1π2 , we have         fx=cos x which being a trigonometrical function is continuous and differentiable in 2nπ, 4n+1π2When, x is in second quadrant or in third quadrant, i.e., 4n+1π2<x<4n+3π2 , we have         fx=-cos x which being a trigonometrical function is continuous and differentiable in 4n+1π2, 4n+3π2When, x is in fourth quadrant, i.e., 4n+3π2< x≤2n+2π , we have         fx=cos x which being a trigonometrical function is continuous and differentiable in 4n+3π2, 2n+2πThus possible point of non-differentiability of fx are x=4n+1π2, 4n+3π2Now, LHD at x=4n+1π2 =limx→4n+1π2- fx- f4n+1π2x-4n+1π2                                               =limx→4n+1π2-  cos x- 0x-4n+1π2                                              =limx→4n+1π2-  -sin x1-0      By L'Hospital rule                                              =-1And  RHD at x=4n+1π2 =limx→4n+1π2+ fx- f4n+1π2x-4n+1π2                                              =limx→4n+1π2+  -cos x- 0x-4n+1π2                                              =limx→4n+1π2+  sin x1-0      By L'Hospital rule                                              =1∴lim x→4n+1π2-fx ≠limx→4n+1π2+fxSo fx is not differentiable at x=4n+1π2Now, LHD at x=4n+3π2 =limx→4n+1π2- fx- f4n+3π2x-4n+3π2                                              =limx→4n+3π2-  -cos x- 0x-4n+3π2                                              =limx→4n+3π2-  sin x1-0      By L'Hospital rule                                              =1And  RHD at x=4n+3π2 =limx→4n+3π2+ fx- f4n+3π2x-4n+3π2                                              =limx→4n+3π2+  cos x- 0x-4n+3π2                                              =limx→4n+3π2+ -sin x1-0      By L'Hospital rule                                              =-1∴lim x→4n+3π2-fx ≠limx→4n+3π2+fxSo fx is not differentiable at x=4n+3π2Therefore, fx is neither differentiable at 4n+1π2 nor at 4n+3π2i.e. fx is not differentiable at odd multiples of π2i.e. fx is not differentiable at x=2n+1π2Therefore, f(x) is everywhere continuous but not differentiable at 2n+1π2 .

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Question 20:

The function fx=sin πx-π4+x2, where [⋅] denotes the greatest integer function, is
(a) continuous as well as differentiable for all x ∈ R
(b) continuous for all x but not differentiable at some x
(c) differentiable for all x but not continuous at some x.
(d) none of these

Answer:


(a) continuous as well as differentiable for all x ∈ R

Here, fx=sin πx-π4+x2
 
Since, we know that πx-π=nπ and sinnπ=0.
    
∵4+x2≠0

∴f(x) = 0 for all x

Thus, f(x) is a constant function and it is continuous and differentiable everywhere.



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Question 21:

Let f (x) = a + b |x| + c |x|4, where a, b, and c are real constants. Then, f (x) is differentiable at x = 0, if
(a) a = 0
(b) b = 0
(c) c = 0
(d) none of these

Answer:

(b) b = 0

We have,fx=a+bx+cx4fx=a+bx+cx4        x≥0a-bx+cx4        x<0Here, fx is differentiable at x=0∴ LHD at x=0 = RHD at x=0⇒limx→0-fx-f0x-0=limx→0+fx-f0x-0⇒limx→0-a-bx+cx4-ax=limx→0+a+bx+cx4-ax⇒limh→0a-b0-h+c0-h4-a0-h=limh→0a+b0+h+c0+h4-a0+h⇒limh→0a+bh+ch4-a-h=limh→0a+bh+ch4-ah⇒limh→0bh+ch4-h=limh→0bh+ch4h⇒limh→0-b-ch3=limh→0b+ch3⇒-b=b⇒2b=0⇒b=0

Page No 9.19:

Question 22:

If f (x) = |3 − x| + (3 + x), where (x) denotes the least integer greater than or equal to x, then f (x) is
(a) continuous and differentiable at x = 3
(b) continuous but not differentiable at x = 3
(c) differentiable nut not continuous at x = 3
(d) neither differentiable nor continuous at x = 3

Answer:

(d) neither differentiable nor continuous at x = 3

We have,fx=3-x+3+x, where x denotes the least integer greater than or equal to x.fx=3-x+3+3,                  2<x<3-3+x+3+4,                  3<x<4⇒ fx=-x+9                2<x<3x+4                    3<x<4Here,LHL at x=3=limx→3-fx=limx→3--x+9=-3+9=6RHL at x=3=limx→3+fx=limx→3-x+4=3+4=7Since, LHL at x=3≠RHL at x=3Hence, given function is not continuous at x= 3Therefore, the function will also not be differentiable at x=3

Page No 9.19:

Question 23:

If fx=11+e1/x,     x≠00,     x=0 then f (x) is

(a) continuous as well as differentiable at x = 0
(b) continuous but not differentiable at x = 0
(c) differentiable but not continuous at x = 0
(d) none of these

Answer:

(d) none of these

we have,

(LHL at x = 0 )=  limx →0- f(x) = limh →0 f(0 - h) = limh →0 f(- h)= limh →0 11 + e1/-h= limh →0 11 + 1e1/h            [limh→01e1/h = 0]  =   11 + 0= 1(RHL at x = 0) =  limx →0+ f(x) = limh →0 f(0 + h)=  limh →0 11 + e1/h=  11 + e1/0  = 11 + e∞ = 11 + ∞     

So, f(x) is not continuous at x = 0

Differentiability at x = 0

(LHD at x = 0 ) = limx →0- f(x) - f(0)x - 0= limh→0 f(0 -h) - f(0)0 -h - 0= limh→0 f(-h) - 0 -h= limh→0  11 + e1/ - h -h= limh→0  11 + 1e1/ h -h         = limh→0  11 + 0 -h  = limh→0  1 -h = -∞(RHD at x = 0) = limx →0+ f(x) - f(0)x - 0= limh →0 f(0 + h) - f(0)0 +h - 0= limh →0 f(h) -0 h = limh→0  11 + e1/  hh = ∞So, f(x) is also not differentiable at x = 0.




 

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Question 24:

If fx=1-cos xx sin x,x≠012             ,x=0

then at x = 0, f (x) is
(a) continuous and differentiable
(b) differentiable but not continuous
(c) continuous but not differentiable
(d) neither continuous nor differentiable

Answer:

(a) continuous and differentiable
 

we have,

fx=1-cos xx sin x,x≠012             ,x=0

fx=1-cos xx sin x,x≠012             ,x=0Continuity at x = 0(LHL at x = 0) = limx →0- f(x)                                               = limh →0 f(0 - h)                               = limh →0 f(- h)                     =  limh →0 1-cos (-h)(-h) sin (-h)                            = limh →0 1-cos hh sin h                       = limh →0 1-cosh   limh →0 1h sin h                       = 1 -cos(0) .  10 sin 0                         = 0                     


(RHL at x = 0) = limx →0+ f(x)                                               = limh →0 f(0 + h)                               = limh →0 f( h)                     =  limh →0 1-cos (h)(h) sin (h)                            = limh →0 1-cos hh sin h                       = limh →0 1-cosh   limh →0 1h sin h                       = 1 - cos 0. 10 sin 0                         = 0


Hence, f(x)is continuous at x = 0.


For differentiability at x = 0

(LHD at x = 0 ) =limx →0- f(x) - f(0)x -0                                        =  limh→0 f(0 - h) - f(0)0 - h -0                           = limh→0 f(- h) -12- h                          = limh→0 1 - cos(-h)- h sin(-h) -12- h                             =  1h limh→0 1 - cosh h sin h  -limh→0 12                             =   12 - 0=  12                                       

RHD at x = 0 ) =limx →0+ f(x) - f(0)x -0                                        =  limh→0 f(0 + h) - f(0)0 - h -0                           = limh→0 f( h) -12- h                          = limh→0 1 - cos(h)- h sin(h) -12- h                          =  -1hlimh→0 1 - cosh h sin h  -limh→0 12                         =  12 - 0 =  12

 

Page No 9.19:

Question 25:

The set of points where the function f (x) given by f (x) = |x − 3| cos x is differentiable, is
(a) R
(b) R − {3}
(c) (0, ∞)
(d) none of these

Answer:

(b) R-3

LHD at x=3=limx→3-fx-f3x-3LHD at x=3=limh→0f3-h-f33-h-3LHD at x=3=limh→0f3-h-f3-hLHD at x=3=limh→03-h-3cos3-h-f3-hLHD at x=3=limh→0hcos3-h-0-h=-cos3RHD at x=3=limx→3+fx-f3x-3RHD at x=3=limh→0f3+h-f33+h-3RHD at x=3=limh→0f3+h-f3hRHD at x=3=limh→03+h-3cos3+h-f3hRHD at x=3=limh→0hcos3+h-0h=cos3

So, f(x) is not differentiable at x = 3.

Also, f(x) is differentiable at all other points because both modulus and cosine functions are differentiable and the product of two differentiable function is differentiable.

Page No 9.19:

Question 26:

Let fx=1  ,x≤-1x,-1<x<10  ,x≥1 Then, f is

(a) continuous at x = − 1
(b) differentiable at x = − 1
(c) everywhere continuous
(d) everywhere differentiable

Answer:

(b) differentiable at x = − 1

fx=1  ,x≤-1x,-1<x<10  ,x≥1

Differentiabilty at x = − 1
(LHD x = − 1)
                                lim x →- 1- f(x) - f(-1) x + 1=lim x →- 1 f(x) - f(-1) x + 1= lim x →- 1 1 - 1 -1 + 1= 0

(RHD x = − 1)
                              = limx →-1+f (x) - f(-1)x + 1 = limx →-1f (x) - f(-1)x + 1 = limx →-1f (x) - f(-1)x + 1= limx →-1|x| - |-1|x + 1= 1 - 1|-1 + 1= 0 

Page No 9.19:

Question 27:

The function f(x) = e|x| is
(a) continuous every where but not differentiable at x = 0
(b) continuous and differentiable everywhere
(c) not continuous at x = 0
(d) none of these

Answer:


The given function is f(x) = e|x|.

We know

If f is continuous on its domain D, then f is also continuous on D.

Now, the identity function p(x) = x is continuous everywhere.

So, g(x) = px=x is also continuous everywhere.

Also, the exponential function ax, a > 0 is continuous everywhere.

So, h(x) = ex is continuous everywhere.

The composition of two continuous functions is continuous everywhere.

∴fx=hogx=ex is continuous everywhere.

Now,

gx=x=x,x≥0-x,x<0

Lg'0=limh→0g0-h-g0-h

⇒Lg'0=limh→0--h-0-h

⇒Lg'0=limh→0h-h

⇒Lg'0=-1

And

Rg'0=limh→0g0+h-g0h

⇒Rg'0=limh→0h-0h

⇒Rg'0=limh→0hh

⇒Rg'0=1

∴Lg'0≠Rg'0

So, gx=x is not differentiable at x = 0.

We know

The exponential function ax, a > 0 is differentiable everywhere.

So, h(x) = ex is differentiable everywhere.

We know that, the composition of differentiable functions is differentiable.
 
Now, ex is differentiable everywhere, but x is not differentiable at x = 0.

∴fx=hogx=ex is differentiable everywhere except at x = 0.   

Thus, the function f(x) = e|x| is continuous every where but not differentiable at x = 0.

Hence, the correct answer is option (a).

Page No 9.19:

Question 28:

The set of points where the function f(x) = |2x – 1| sin x is differentiable, is

(a) R

(b) R-12

(c) (0, ∞)

(d) none of these

Answer:


Let gx=2x-1 and hx=sinx.

We know that, the trigonometric functions are differentiable in their respective domain.

So, hx=sinx is differentiable for all x ∈ R.

Now,

gx=2x-1=2x-1,x≥12-2x-1,x<12

(2x − 1) and −(2x − 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all x<12 and for all x>12.

So, we need to check the differentiability of g(x) at x=12.

We have

Lg'12=limh→0g12-h-g12-h

⇒Lg'12=limh→0-212-h-1-2×12-1-h

⇒Lg'12=limh→02h-h

⇒Lg'12=-2

And

Rg'12=limh→0g12+h-g12h

⇒Rg'12=limh→0212+h-1-2×12-1h

⇒Rg'12=limh→02hh

⇒Rg'12=2

∴Lg'12≠Rg'12

So, gx=2x-1 is not differentiable at x=12.

The function gx=2x-1 is differentiable for all x∈R-12.

We know that, the product of two differentiable functions is differentiable.

∴fx=gx×hx=2x-1sinx is differentiable for all x∈R-12.

Thus, the set of points where the function fx=2x-1sinx is differentiable is R-12.

Hence, the correct answer is option (b).



Page No 9.20:

Question 1:

The function f(x) = |x + 1| is not differentiable at x = ____________.

Answer:


The given function is fx=x+1.

fx=x+1=x+1,x≥-1-x+1,x<-1

Now, (x + 1) and −(x + 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all x>-1and for all x<-1.

So, we need to check the differentiability of f(x) at x=-1.

We have,

Lf'-1=limh→0f-1-h-f-1-h

⇒Lf'-1=limh→0--1-h+1--1+1-h

⇒Lf'-1=limh→0h-h

⇒Lf'-1=-1

And

Rf'-1=limh→0f-1+h-f-1h

⇒Rf'-1=limh→0-1+h+1--1+1h

⇒Rf'-1=limh→0hh

⇒Rf'-1=1

∴Lf'-1≠Rf'-1

So, f(x) is not differentiable at x=-1.

Thus, the function f(x) = |x + 1| is not differentiable at x = −1.


The function f(x) = |x + 1| is not differentiable at x = ___−1___.

Page No 9.20:

Question 2:

The function g(x) = |x – 1| + |x + 1| is not differentiable at x = ____________.

Answer:


x-1=x-1,x≥1-x-1,x<1

x+1=x+1,x≥-1-x+1,x<-1

∴gx=x-1+x+1=-x-1-x+1,x<-1-x-1+x+1,-1≤x<1x-1+x+1,x≥1

⇒gx=x-1+x+1=-2x,x<-12,-1≤x<12x,x≥1

When x < −1, g(x) = −2x which being a polynomial function is continuous and differentiable.

When −1 ≤ x < 1, g(x) = 2 which being a constant function is continuous and differentiable.

When x ≥ 1, g(x) = 2x which being a polynomial function is continuous and differentiable.

Thus, the possible points of non-differentiability of g(x) are x = −1 and x = 1.

Now,

 Lg'-1=limh→0g-1-h-g-1-h

⇒Lg'-1=limh→0-2-1-h-2-h

⇒Lg'-1=limh→02h-h

⇒Lg'-1=-2

And

Rg'-1=limh→0g-1+h-g-1h

⇒Rg'-1=limh→02-2-h

⇒Rg'-1=limh→00-h

⇒Rg'-1=0

∴Lg'-1≠Rg'-1

So, g(x) is not differentiable at x = −1.

Also,

 Lg'1=limh→0g1-h-g1-h

⇒Lg'1=limh→02-2×1-h

⇒Lg'1=limh→00-h

⇒Lg'1=0

And

Rg'1=limh→0g1+h-g1h

⇒Rg'1=limh→021+h-2×1h

⇒Rg'1=limh→02hh

⇒Rg'1=2

∴Lg'1≠Rg'1

So, g(x) is not differentiable at x = 1.

Thus, the function gx=x-1+x+1 is not differentiable at x = −1 and x = 1.


The function g(x) = |x – 1| + |x + 1| is not differentiable at x = ___±1___.

Page No 9.20:

Question 3:

The set of points where f(x) = x – [x] not differentiable is ____________.

Answer:


Let g(x) = x and h(x) = [x].

Every polynomial function is differentiable for all x ∈ R. So, g(x) = x is differentiable for all x ∈ R.

Also, the function h(x) = [x] is discontinuous at all integral values of x i.e. x ∈ Z. So, h(x) = [x] is not differentiable at all integral values of x i.e. x ∈ Z.

Now, f(x) = g(x) − h(x) = x − [x]

So, the function f(x) = x − [x] is differentiable for all x ∈ R except at all integral values of x i.e. x ∈ Z. The function f(x) = x − [x] is not differentiable for all x ∈ R − Z. 

Thus, the set of points where f(x) = x – [x] not differentiable is R − Z.


The set of points where f(x) = x – [x] not differentiable is ___R − Z___.

Page No 9.20:

Question 4:

The number of points in [–π, π] where f(x) = sin–1 (sin x) is not differentiable is. ____________.

Answer:



fx=sin-1sinx=-x-π,-π≤x≤-π2x,-π2≤x≤π2π-x,π2≤x≤π

Let us check the differentiability of the function at x=-π2 and x=π2.

At x=-π2,

Lf'-π2=limx→-π2-fx-f-π2x--π2⇒Lf'-π2=limx→-π2-x-π--π2x+π2⇒Lf'-π2=limx→-π2-x+π2x+π2⇒Lf'-π2=-1

Rf'-π2=limx→-π2+fx-f-π2x--π2⇒Rf'-π2=limx→-π2x--π2x+π2⇒Rf'-π2=limx→-π2x+π2x+π2⇒Rf'-π2=1

∴Lf'-π2≠Rf'-π2

So, the function f(x) is not differentiable at x=-π2.

At x=π2,

Lf'π2=limx→π2-fx-fπ2x-π2⇒Lf'π2=limx→π2x-π2x-π2⇒Lf'π2=1

Rf'π2=limx→π2+fx-fπ2x-π2⇒Rf'π2=limx→π2π-x-π2x-π2⇒Rf'π2=limx→π2-x-π2x-π2⇒Rf'π2=-1

∴Lf'π2≠Rf'π2

So, the function f(x) is not differentiable at x=π2.

Thus, the function f(x) = sin–1(sinx), x ∈ [–π, π] is not differentiable at x=-π2 and x=π2.


 The number of points in [–π, π] where f(x) = sin–1 (sin x) is not differentiable is       2      .

Page No 9.20:

Question 5:

The function f(x) = cos–1(cos x), x ∈ (–2π, 2π) is not differentiable at x = ____________.

Answer:


fx=cos-1cosx=x+2π-2π≤x≤-π-x,-π≤x≤0x,0≤x≤π2π-x,π≤x≤2π

Let us check the differentiability of the function at x=-π, x = 0 and x=π.

At x=-π,

Lf'-π=limx→-π-fx-f-πx--π⇒Lf'-π=limx→-πx+2π-πx+π⇒Lf'-π=limx→-πx+πx+π⇒Lf'-π=1

Rf'-π=limx→-π+fx-f-πx--π⇒Rf'-π=limx→-π-x-πx+π⇒Rf'-π=limx→-π-x+πx+π⇒Rf'-π=-1

∴Lf'-π≠Rf'-π

So, the function f(x) is not differentiable at x=-π.

At x=0,

Lf'0=limx→0-fx-f0x-0⇒Lf'0=limx→0-x-0x⇒Lf'0=limx→0-xx⇒Lf'0=-1

Rf'0=limx→0+fx-f0x-0⇒Rf'0=limx→0x-0x-0⇒Rf'0=limx→0xx⇒Rf'0=1

∴Lf'0≠Rf'0

So, the function f(x) is not differentiable at x=0.

At x=π,

Lf'π=limx→π-fx-fπx-π⇒Lf'π=limx→-πx-πx-π⇒Lf'π=1

Rf'π=limx→π+fx-fπx-π⇒Rf'π=limx→π2π-x-πx-π⇒Rf'π=limx→π-x-πx-π⇒Rf'π=-1

∴Lf'π≠Rf'π

So, the function f(x) is not differentiable at x=π.

Thus, the function f(x) = cos–1(cos x), x ∈ (–2π, 2π) is not differentiable at x=-π, x = 0 and x=π.


The function f(x) = cos–1(cos x), x ∈ (–2π, 2π) is not differentiable at x =       -π,0,π      .

Page No 9.20:

Question 6:

The function fx=sin x, -π2, π2 is not differentiable at x = ____________.

Answer:


We know that, x is not differentiable at x = 0.

Therefore, sinx is not differentiable when sinx=0.

sinx=0

⇒x=nπ, n ∈ Z

⇒x= ...,-2π,-π,0,π,2π,...

Now, the only value of x lying in given interval -π2, π2 at which the function fx=sinx is not differentiable is 0.

Thus, the function fx=sinx, x∈-π2, π2 is not differentiable at x = 0.


The function fx=sin x, -π2, π2 is not differentiable at x = ___0___.

Page No 9.20:

Question 7:

Let fx=ax2+3,x>1x+52,x≤1 . If f(x) is differentiable at x = 1, then a = ____________.

Answer:


The given function fx=ax2+3,x>1x+52,x≤1 is differentiable at x = 1.

∴Lf'1=Rf'1

⇒limh→0f1-h-f1-h=limh→0f1+h-f1h

⇒limh→01-h+52-1+52-h=limh→0a1+h2+3-1+52h

⇒limh→0-h-h=limh→0a1+h2-12h

⇒1=limh→0a-12+a2h+h2h

⇒1=limh→0a-12+ah2+hh

⇒1=limh→0a-12h+limh→0a2+h

Here, LHS is finite.

So, for RHS to be finite, we must have

a-12=0

⇒a=12

Thus, the value of a is 12.


Let fx=ax2+3,x>1x+52,x≤1 . If f(x) is differentiable at x = 1, then a =      12     .

Page No 9.20:

Question 8:

If f(x) = x |x|, then f' (–1) = ____________.

Answer:


x=x,x≥0-x,x<0

∴fx=xx=x2,x≥0-x2,x<0

Now,

Lf'-1=limh→0f-1-h-f-1-h

⇒Lf'-1=limh→0--1-h2---12-h

⇒Lf'-1=limh→0-1+2h+h2+1-h

⇒Lf'-1=limh→0-2h+h2-h

⇒Lf'-1=limh→02+h

⇒Lf'-1=2+0=2

Also,

Rf'-1=limh→0f-1+h-f-1h

⇒Rf'-1=limh→0--1+h2---12h

⇒Rf'-1=limh→0-1-2h+h2+1h

⇒Rf'-1=limh→0--2h+h2h

⇒Rf'-1=limh→02-h

⇒Rf'-1=2-0=2

So, Lf'-1=Rf'-1=2

∴f'-1=2


If f(x) = x|x|, then f'(–1) = ___2___.

Page No 9.20:

Question 9:

If f(x) = x |x|, then f' (2) =____________.

Answer:


x=x,x≥0-x,x<0

∴fx=xx=x2,x≥0-x2,x<0

Now,

Lf'2=limh→0f2-h-f2-h

⇒Lf'2=limh→02-h2-22-h

⇒Lf'2=limh→04-4h+h2-4-h

⇒Lf'2=limh→0-4+hh-h

⇒Lf'2=limh→04-h

⇒Lf'2=4-0=4

Also,

Rf'2=limh→0f2+h-f2h

⇒Rf'2=limh→02+h2-22h

⇒Rf'2=limh→04+4h+h2-4h

⇒Rf'2=limh→04+hhh

⇒Rf'2=limh→04+h

⇒Rf'2=4+0=4

So, Lf'2=Rf'2=4.

∴f'2=4


If f(x) = x|x|, then f'(2) = ___4____.

Page No 9.20:

Question 10:

The set of point where the function f(x) = |2x – 1| is differentiable, is ____________.

Answer:


The given function is fx=2x-1.

fx=2x-1=2x-1,x≥12-2x-1,x<12

Now, (2x − 1) and −(2x − 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all x>12and for all x<12.

So, we need to check the differentiability of f(x) at x=12.

We have,
Lf'12=limh→0f12-h-f12-h

⇒Lf'12=limh→01-212-h-2×12-1-h

⇒Lf'12=limh→02h-h

⇒Lf'12=-2

And
Rf'12=limh→0f12+h-f12h

⇒Rf'12=limh→0212+h-1-2×12-1h

⇒Rf'12=limh→02hh

⇒Rf'12=2

∴Lf'12≠Rf'12

So, f(x) is not differentiable at x=12.

Thus, the set of points where the function f(x) = |2x – 1| is differentiable is R-12.


The set of point where the function f(x) = |2x – 1| is differentiable, is      R-12     .

Page No 9.20:

Question 11:

The set of points where the function f(x)=x+1,x<22x-1,x≥2is not differentiable, is ____________.

Answer:


The given function is fx=x+1,x<22x-1,x≥2.

(x + 1) and (2x − 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all x < 2 and for all x > 2.

So, we need to check the differentiability of f(x) at x = 2.

We have

Lf'2=limh→0f2-h-f2-h

⇒Lf'2=limh→02-h+1-2×2-1-h

⇒Lf'2=limh→03-h-3-h

⇒Lf'2=limh→0-h-h

⇒Lf'2=1

And

Rf'2=limh→0f2+h-f2h

⇒Rf'2=limh→022+h-1-2×2-1h

⇒Rf'2=limh→03+2h-3h

⇒Rf'2=limh→02hh

⇒Rf'2=2

∴Lf'2≠Rf'2

So, f(x) is not differentiable at x = 2.

Thus, the set of points where the function f(x) is not differentiable is {2}.


The set of points where the function fx=x+1,x<22x-1,x≥2is not differentiable, is _____{2}______.

Page No 9.20:

Question 12:

An example of a function which is everywhere continuous but fails to be differentiable exactly at two points is ____________.

Answer:


Consider the function gx=x-1+x+1.

x-1=x-1,x≥1-x-1,x<1

x+1=x+1,x≥-1-x+1,x<-1

∴gx=x-1+x+1=-x-1-x+1,x<-1-x-1+x+1,-1≤x<1x-1+x+1,x≥1

⇒gx=x-1+x+1=-2x,x<-12,-1≤x<12x,x≥1

When x < −1, g(x) = −2x which being a polynomial function is continuous and differentiable.

When −1 ≤ x < 1, g(x) = 2 which being a constant function is continuous and differentiable.

When x ≥ 1, g(x) = 2x which being a polynomial function is continuous and differentiable.

Let us check the continuity and differentiability of g(x) at x = −1 and x = 1.

At x = −1,

LHL = limx→-1-gx=limx→-1-2x=-2×-1=2

RHL = limx→-1+gx=limx→-12=2

g-1=2

Since limx→-1-gx=limx→-1+gx=g-1, so the function g(x) is continuous at x = −1.

At x = 1,

LHL = limx→1-gx=limx→12=2

RHL = limx→1+gx=limx→12x=2×1=2

g1=2×1=2

Since limx→1-gx=limx→1+gx=g1, so the function g(x) is continuous at x = 1.

Thus, the function g(x) is continuous everywhere i.e. for all x ∈ R.

Now,

 Lg'-1=limh→0g-1-h-g-1-h

⇒Lg'-1=limh→0-2-1-h-2-h

⇒Lg'-1=limh→02h-h

⇒Lg'-1=-2

And

Rg'-1=limh→0g-1+h-g-1h

⇒Rg'-1=limh→02-2-h

⇒Rg'-1=limh→00-h

⇒Rg'-1=0

∴Lg'-1≠Rg'-1

So, g(x) is not differentiable at x = −1.

Also,

 Lg'1=limh→0g1-h-g1-h

⇒Lg'1=limh→02-2×1-h

⇒Lg'1=limh→00-h

⇒Lg'1=0

And

Rg'1=limh→0g1+h-g1h

⇒Rg'1=limh→021+h-2×1h

⇒Rg'1=limh→02hh

⇒Rg'1=2

∴Lg'1≠Rg'1

So, g(x) is not differentiable at x = 1.

Thus, the function g(x) is differentiable everywhere except at x = −1 and x = 1.


An example of a function which is everywhere continuous but fails to be differentiable exactly at two points is      gx=x-1+x+1     .

Page No 9.20:

Question 13:

The set of points where f(x) = cos |x| is differentiable, is ____________.

Answer:


We know

x=x,x≥0-x,x<0

∴fx=cosx=cosx,x≥0cos-x,x<0

⇒fx=cosx=cosx,x≥0cosx,x<0                  cos-θ=cosθ

We know that, cosine function is differentiable in its domain. So, f(x) is differentiable for all x < 0 and x > 0.

Let us check the differentiability of fx=cosxat x = 0.

Now,

Lf'0=limh→0f0-h-f0-h

⇒Lf'0=limh→0cos-h-cos0-h

⇒Lf'0=limh→0cosh-1-h

⇒Lf'0=limh→0-2sin2h2-h

⇒Lf'0=limh→0sinh2h2×limh→0sinh2

⇒Lf'0=1×0

⇒Lf'0=0

And

Rf'0=limh→0f0+h-f0h

⇒Rf'0=limh→0cosh-cos0h

⇒Rf'0=limh→0cosh-1h

⇒Rf'0=limh→0-2sin2h2h

⇒Rf'0=-limh→0sinh2h2×limh→0sinh2

⇒Rf'0=-1×0

⇒Rf'0=0

∴Lf'0=Rf'0

So, f(x) is differentiable at x = 0. Thus, the function f(x) is differentiable everywhere.

Hence, the set of points where fx=cosxis differentiable is R (set of real real numbers).


The set of points where f(x) = cos |x| is differentiable, is _____R_____.

Page No 9.20:

Question 14:

The set of points where f(x) = |sin x| is not differentiable, is ____________.

Answer:


Let gx=x=x,x≥0-x,x<0

Now,

Lg'0=limh→0g0-h-g0-h

⇒Lg'0=limh→0--h-0-h

⇒Lg'0=limh→0h-h

⇒Lg'0=-1

And

Rg'0=limh→0g0+h-g0h

⇒Rg'0=limh→0h-0h

⇒Rg'0=limh→0hh

⇒Rg'0=1

∴Lg'0≠Rg'0

So, x is not differentiable at x = 0.

Therefore, fx=sinx is not differentiable when sinx=0.

sinx=0

⇒x=nπ, n∈Z

Thus, the set of points where fx=sinx is not differentiable is x=nπ:n∈Z.


The set of points where f(x) = |sin x| is not differentiable, is      x=nπ:n∈Z     .

Page No 9.20:

Question 15:

The set of points at which the function fx=1logxis not differentiable, is ____________.

Answer:


The given function is fx=1logx.

For f(x) to be defined,

x≠0 and logx≠0

⇒x≠0 and x≠1

⇒x≠0 and x≠±1

Thus, the function f(x) is not defined when x = −1, x = 0 and x = 1.

We know that, the logarithmic function is differentiable at each point in its domain. Every constant function is differentiable at each x ∈ R. Also, the quotient of two differentiable functions is differentiable.

So, the function fx=1logx is not differentiable at x = −1, x = 0 and x = 1.

Thus, the set of points at which the function fx=1logxis not differentiable is {−1, 0, 1}.


The set of points at which the function fx=1logxis not differentiable, is ___{−1, 0, 1}___.  

Page No 9.20:

Question 1:

Define differentiability of a function at a point.

Answer:

Let f(x) be a real valued function defined on an open interval (a,b) and let c∈(a,b).
Then f(x) is said to be differentiable or derivable at x=c iff
limx→c f(x) - f(c)x-c exists finitely.

or,  f'(c) =limx→c f(x) - f(c)x-c.

Page No 9.20:

Question 2:

Is every differentiable function continuous?

Answer:

Yes, if a function is differentiable at a point then it is necessary continuous at that point.

Proof :Let a function f(x) be differentiable at x=c . Then, limx→c f(x)-f(c)x-cexists finitely.Let limx→c f(x)-f(c)x-c=f'(c)In order to prove that f(x) is continous at x=c , it is sufficient to show that  limx→c f(x)=f(c) limx→c f(x)=limx→c f(x)-f(c)x-cx-c+f(c)⇒ limx→c f(x)=limx→c f(x)-f(c)x-cx-c+f(c)⇒ limx→c f(x)=limx→c f(x)-f(c)x-c. lim  x→c x-c+f(c)⇒ limx→c f(x)=f'(c) ×0+f(c)⇒ limx→c f(x)=f(c)Hence, f(x) is continuous at x=c.

Page No 9.20:

Question 3:

Is every continuous function differentiable?

Answer:

No, function may be continuous at a point but may not be differentiable at that point .
For example: function  f(x) = |x| is continuous at x=0 but it is not differentiable at x=0.

Page No 9.20:

Question 4:

Give an example of a function which is continuos but not differentiable at at a point.

Answer:

Consider a function, f(x) = x,        x>0-x,     x≤0
This mod function is continuous at x=0 but not differentiable at x=0.

Continuity at x=0, we have:

(LHL at x = 0)

 limx→0- f(x) = limh→0 f(0-h) = limh→0 -(0-h) = 0

(RHL at x = 0)

 limx→0+ f(x) = limh→0 f(0+h) = limh→0 (0+h) = 0

and  f(0) = 0

Thus, limx→0- f(x) = limx→0+ f(x) = f(0).

Hence, f(x) is continuous at x=0.

Now, we will check the differentiability at x=0, we have:

(LHD at x = 0)

limx→0- f(x) - f(0)x-0 = limh→0 f(0-h) - f(0)0-h-0 = limh→0 -(0-h) - 0-h = -1

(RHD at x = 0)

limx→0+ f(x) - f(0)x-0 = limh→0 f(0+h) - f(0)0+h-0 = limh→0 0+h - 0h = 1

Thus,  limx→0- f(x)  ≠ limx→0+ f(x) 

Hence f(x) is not differentiable at x=0.

Page No 9.20:

Question 5:

If f (x) is differentiable at x = c, then write the value of limx→cf x.

Answer:

Given: f(x) is differentiable at x=c. Then,
limx→c f(x) - f(c)x-c exists finitely.

or,  limx→c f(x) - f(c)x-c = f'(c).

Consider,
  limx→c f(x) = limx→c f(x) - f(c)x-c (x-c) + f(c) limx→c f(x) = limx→c f(x) - f(c)x-c (x-c) + f(c)limx→c f(x) = limx→c f(x) - f(c)x-c limx→c (x-c) + f(c)limx→c f(x) = f'(c) × 0 + f(c) = f(c)
       

Page No 9.20:

Question 6:

If f (x) = |x − 2| write whether f' (2) exists or not.

Answer:

Given: f(x) = x-2 = x-2,     x>2-x+2,  x≤2

Now,

(LHD at x = 2)

 limx→2- f(x) - f(2)x-2 = limh→0 f(2-h) - f(2)2-h-2 = limh→0 (-2+h+2) -0-h = -1

(RHD at x = 2)

 limx→2+f(x) - f(2)x-2 = limh→0 f(2+h) - f(2)2+h-2 = limh→0 2+h-2 - 0h=1

Thus, (LHD at x = 2) ≠ (RHD at x = 2)

Hence, limx→2 f(x) - f(2)x-2 = f'(2) does not exist.

Page No 9.20:

Question 7:

Write the points where f (x) = |loge x| is not differentiable.

Answer:

Given: f(x) = logex = -logex,    0<x<1logex,        x≥1
Clearly f(x) is differentiable for all x>1 and for all x<1. So, we have to check the differentiability at x=1.
We have,
(LHD at x = 1) 

   limx→1- f(x) - f(1)x-1 

 = limx→1- -log x - log 1x-1=- limx→1-  log xx-1=- limh→0 log (1-h)1-h-1=-limh→0log (1-h)-h =-1

(RHD at x=1)
= limx→1+ f(x) - f(1)x-1
      
=limx→1+ log x - log 1x-1=limx→1+ log xx-1= limh→0  log (1+h)1+h-1= limh→0 log (1+h)h=1

Thus, (LHD at x =1) ≠ (RHD at x =1)
So, f(x) is not differentiable at x=1.

Page No 9.20:

Question 8:

Write the points of non-differentiability of f x=log x.

Answer:

We have,
f (x) = |log |x||

x=-x        -∞<x<-1-x        -1<x<0  x        0<x<1  x        1<x<∞log x=log -x        -∞<x<-1log -x        -1<x<0  log x        0<x<1 log x        1<x<∞log x=log -x        -∞<x<-1-log -x        -1<x<0 -log x        0<x<1 log x        1<x<∞
LHD at x=-1=limx→-1-fx-f-1x+1                             =limx→-1-log -x-0x+1                             =limh→0log 1+h-1-h+1                             =-limh→0log 1+hh=-1

RHD at x=-1=limx→-1+fx-f-1x+1                             =limx→-1+-log -x-0x+1                             =limh→0-log 1-h-1+h+1                             =limh→0-log 1-hh=1
Here, LHD ≠ RHD
So, function is not differentiable at x = − 1

At 0 function is not defined.
LHD at x=1=limx→1-fx-f1x-1                        =limx→1--log x-0x-1                       =limh→0-log 1-h1-h-1                       =-limh→0log 1-hh=-1
RHD at x=1=limx→1+fx-f1x-1                        =limx→1+log x-0x-1                       =limh→0log 1+h1+h-1                       =limh→0log 1+hh=1
Here, LHD ≠ RHD
So, function is not differentiable at x = 1
Hence, function is not differentiable at x = 0 and ± 1

Page No 9.20:

Question 9:

Write the derivative of f (x) = |x|3 at x = 0.

Answer:

Given: f(x) = x3 = x3,      x≥0-x3,   x<0

(LHD at x = 0)
limx→0-f(x) - f(0)x-0= limh→0 f(0-h) - f(0) x= limh→0 h3-h = 0.

(RHD at x = 0)

 limx→0+ f(x) - f(0)x-0 = limx→0+ f(0+h) - f(0)x= limh→0 h3 - 0h = 0

and f(0) = 0.

Thus, (LHD at x=0) = (RHD at x = 0) = f(0)

Hence, limx→0f(x) - f(0)x-0 = f'(0) = 0

Page No 9.20:

Question 10:

Write the number of points where f (x) = |x| + |x − 1| is continuous but not differentiable.

Answer:

Given:
         f(x) = x + x-1        
⇒f(x) =  -x-(x-1),    x<0x-(x-1),      0≤x<1x+(x-1),        x≥1⇒ f(x) =  -2x+1,        x<01,                 0≤x<12x-1,           x≥1

When x<0, we have:

f(x) =-2x+1 which, being a polynomial function is continuous and differentiable.

When 0≤x<1, we have:
f(x) =1 which, being a constant function is continuous and differentiable on (0,1).

When x≥1, we have:
f(x) =2x-1 which, being a polynomial function is continuous and differentiable on x>2.

Thus, the possible points of non- differentiability of f(x) are 0 and 1.
Now,
(LHD at x = 0)

 limx→0-f(x) - f(0)x-0
                    
 = limx→0 -2x+1 - 1x-0                                  [∵ f(x) =-2x+1,  x<0]

 = limx→0 -2xx =-2

(RHD at x = 0)

= limx→0+ f(x) - f(0)x-0

 = limx→01-1x-1=0                                             [∵ f(x)=1,     0≤x<1 ]

Thus, (LHD at x=0) ≠ (RHD at x=0)

Hence f(x)  is not differentiable at x=0

Now, f(x) is not differentiable at x=1.

(LHD at x = 1)

limx→1-f(x) - f(1)x-1
= limx→1 1- 1x-1  =0

(RHD at x = 1)
 = limx→1+f(x) - f(1)x-1
= limx→1 2x-1 - 1x-1= limx→1 2(x-1)x-1 = 2
 
Thus, (LHD at x =1) ≠ (RHD at x=1)
.
Hence f(x) is not differentiable at x=1.

Therefore, 0,1 are the points where f(x) is continuous but not differentiable.



Page No 9.21:

Question 11:

If limx→cfx-fcx-c exists finitely, write the value of limx→cfx.

Answer:

Given:  limx→cf(x) - f(c)x-c exists finitely. Then,    
 
  limx→c f(x) - f(c)x-c = f'(c).

Now,    
limx→c f(x) = lim x→c f(x) - f(c)x-c (x-c) + f(c)                = limx→c f(x) - f(c)x-c (x-c) + f(c)                =  limx→c f(x) - f(c)x-c limx→c (x-c) + f(c)                = f'(c)×0 + f(c)                 = f(c) 

Page No 9.21:

Question 12:

Write the value of the derivative of f (x) = |x − 1| + |x − 3| at x = 2.

Answer:

Given: f(x) = x-1 + x-3

⇒ f(x) = -(x-1) -(x-3),      x<1x-1 - (x-3),         1≤x<3          (x-1) + (x-3),        x≥3

⇒ f(x) =  -2x+4,     x<12,                1≤x<32x-4,        x≥3

We check differentiability at x = 2

(LHD at x = 2)

  limx→2- f(x) - f(2)x-2 = limh→0 f(2-h) - f(2)2-h-2 = limh→0 2-2-h =0

Page No 9.21:

Question 13:

If f x=x2+9, write the value of limx→4 fx-f4x-4.

Answer:

Given: f(x) = x2 + 9
Now,
       f(4) = 16+9        = 25        =5 

So, f(x) - f(4)x-4 = x2 + 9 - 5x-4

On rationalising the numerator, we get

f(x) - f(4)x-4 = x2+9 -5x-4×x2+9 + 5x2+9 + 5                       = x2+9 - 25(x-4) x2+9 + 5                       = x2 - 16(x-4) x2+9 + 5                      = (x+4)x2+9 + 5

Taking limit x→4, we have

limx→4 f(x) - f(4)x-4 = limx→4 (x+4)x2+9 + 5                                = 810                                = 45



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